25. K 个一组翻转链表 Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

Follow up:

Could you solve the problem in O(1) extra memory space?
You may not alter the values in the list's nodes, only nodes itself may be changed.

Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

方法：

链表分区为已翻转部分+待翻转部分+未翻转部分
每次翻转前，确定翻转链表的范围，通过 k 此循环来确定
记录翻转链表前驱和后继，方便翻转完成后把已翻转部分和未翻转部分连接起来
初始需要两个变量 pre 和 end，pre 代表待翻转链表的前驱，end 代表待翻转链表的末尾
经过k此循环，end 到达末尾，记录待翻转链表的后继 next = end.next
翻转链表，然后将三部分链表连接起来，然后重置 pre 和 end 指针，然后进入下一次循环
特殊情况，当翻转部分长度不足 k 时，在定位 end 完成后，end==null，已经到达末尾，说明题目已完成，直接返回即可

public ListNode reverseKGroup(ListNode head, int k) {
    ListNode dummy = new ListNode(0);
    dummy.next = head;

    ListNode pre = dummy;
    ListNode end = dummy;

    while (end.next != null) {
        for (int i = 0; i < k && end != null; i++) 
             end = end.next;
        if (end == null) break;
        ListNode start = pre.next;
        ListNode next = end.next;
        end.next = null;
        pre.next = reverse(start);
        start.next = next;
        pre = start;
        end = pre;
    }
    return dummy.next;
}

private ListNode reverse(ListNode head) {
    ListNode pre = null;
    ListNode curr = head;
    while (curr != null) {
        ListNode next = curr.next;
        curr.next = pre;
        pre = curr;
        curr = next;
    }